Can someone check whether my answers are okay to the following Judson question?

(a) Let $$f(x)=a_{n}x^{n}+\cdots +a_{m}x^{m}+\cdots a_{1}x+a_{0}$$ and $$g(x)=b_{m}x^{m}+\cdots +b_{1}x+b_{0}$$. Assume that $$deg(f(x))>deg(g(x))$$.

$$f'(x) = na_{n}x^{n-1}+\cdots ma_{m}x^{m-1}+ a_{1}$$

$$g'(x) = mb_{m}x^{m-1}+\cdots + b_{1}$$

$$f'(x)+g'(x)=na_{n}x^{n-1}\cdots + m(a_{m}+b_{m})x^{m-1}+\cdots + (a_{1}+b_{1})$$

$$f(x) + g(x)=a_{n}x^{n}+\cdots +(a_{m}+b_{m})x^{m}+\cdots (a_{1}+b_{1})x+(a_{0}+b_{0})$$

$$(f(x) + g(x))'=na_{n}x^{n-1}+\cdots +m(a_{m}+b_{m})x^{m-1}+\cdots (a_{1}+b_{1})$$

Therefore, $$(f(x)+g(x))'=f'(x)+g'(x)$$.

Define $$D:F[x]\to F[x]$$ by $$D(f(x))=f'(x)$$. Then $$D$$ is operation-preserving. If $$f(x)=g(x)$$, then they must have the same derivative, and so $$D$$ is well-defined. Therefore, $$D$$ is a group homomorphism.

(b) Let $$char(F)=0$$. Then there does not exist a positive integer $$n$$ such that $$nr=0$$ for all $$r\in F$$. If $$ker(D)=\left \{f(x)\in F[x]:f'(x)=na_{n}x^{n-1}+\cdots ma_{m}x^{m-1}+ a_{1}=0 \right \}$$, then $$na_{n}=0$$, ..., $$ma_{m}=0$$, ..., $$(1)a_{1}=0$$. Since $$char(F)=0$$, it must be that $$a_{n}=\cdots =a_{m}=\cdots =a_{1}=0$$. Then $$ker(D)=\left \{f(x)=a_{0}\in F[x] \right \}=F$$.

(c) Let $$char(F)=p$$. Then for any $$r\in F$$, $$pr=0$$. Then any coefficients $$a_{k}$$ such that $$p$$ divides $$a_{k}$$ will make the entire term zero. Then for the coefficients of the derivative to be zero, either the coefficients are already a multiple of $$p$$ or the exponent of $$x$$ is a multiple of $$p$$. Then $$ker(D)=\left \{f(x)\in F[x]:f'(x)=0 \right \}=\left \{f(x)=ax^{pk}\in F[x]:k\in \mathbb{Z},a\in F \right \}$$.

(d) Is there a faster way to prove this without having to define two polynomials, multiply them and find their derivative, then calculate the right side of the equation and compare results?

(e) Assume that $$f(x)$$ has no repeated factors. Then $$f'(x)$$ will be a sum of products with no common factor, since if $$f(x)=g(x)h(x)$$, then $$f'(x)=g(x)h'(x) + h(x)g'(x)$$. Since h(x) and g(x) have no common factors, calculating $$h'(x)$$ and $$g'(x)$$ will also produce factors that are not in common. Therefore, there exists no common factor between $$f(x)$$ and $$f'(x)$$ and so they are relatively prime.

Assume that $$f(x)$$ has a repeated root, where $$f(x)=(x-r)^{2}g(x)$$. Then $$(x-r)$$ divides $$f(x)$$. Then $$f'(x) = (x-r)^{2}g'(x) + g(x)(2(x-r))=(x-r)[(x-r)g'(x)+2g(x)]$$. Then $$(x-r)$$ divides $$f'(x)$$. Then $$f(x)$$ and $$f'(x)$$ are not relatively prime.